Bag 1 contains 4 white balls and 5 black balls, and Bag 2 contains n white balls and 3 black balls. One ball is drawn randomly from Bag 1 and transferred to Bag 2. A ball is then drawn randomly from Bag 2. If the probability, that the ball drawn is white, is $\frac{29}{45}$, then n is equal to:
Let $S = \mathbb{N} \cup \{0\}$. Define a relation R from S to $\mathbb{R}$ by: $R = \{(x, y) : \log_e y = x \log_e (\frac{2}{5}), x \in S, y \in \mathbb{R}\}$. Then, the sum of all the elements in the range of $R$ is equal to:
(A)
$\frac{3}{2}$
(B)
$\frac{10}{9}$
(C)
$\frac{5}{2}$
(D)
$\frac{5}{3}$
MEDIUM
Correct Answer:D
Explanation
Given $S = \{0, 1, 2, 3, ...\}$. Also, $\log_e y = x \log_e (\frac{2}{5})$.
This implies $y = (\frac{2}{5})^x$.
Since $x \in S$, $x$ can take values $0, 1, 2, 3, ...$.
The required sum is $1 + (\frac{2}{5})^1 + (\frac{2}{5})^2 + (\frac{2}{5})^3 + ... = \frac{1}{1 - \frac{2}{5}} = \frac{1}{\frac{3}{5}} = \frac{5}{3}$.